Question

Show that the function f(x) == 1/x is uniformly continuous on the set A = 18,00), where esa
positive constant.​

Answer 1

Answer:

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Answer 2

Answer:

please check my proof

Let ϵ>0δ>0 and a is arbitary positive constant in interval

|x−a|<δ↔|f(x)−f(a)|<ϵ

↔|1x−1a|<ϵ

↔|a−xxa|<ϵ

↔1a|a−xx|<ϵ

|a−xx|<aϵ

choose δ=aϵ

then|1x−1a|<δ↔|1x−1a<aϵa=ϵ

Step-by-step explanation:

More generally:

(1). For a,b∈R, if f:[a,b]→R is continuous then f is uniformly continuous on [a,b].

(2). Corollary. If f:[a,∞) is continuous and limx→∞f(x)=0 then f is uniformly continuous on [a,∞).

Proof of Corollary. Given e>0, take c≥a such that ∀x∈[c,∞)(|f(x)|<e/2).

Let b=c+1.

Take d∈(0,1) such that ∀x,y∈[a,b](|x−y|<d⟹|f(x)−f(y)|<e/2).

Now for all x,y∈(b,∞) we have x,y∈[c,∞) so|f(x)−f(y)|≤|f(x)|+|f(y)|<e/2+e/2=e.

And for a≤x≤b<y, if |x−y|<d then

y>x>y−d>b−d=c+1−d>c,

so {x,y}⊂[c,∞), so |f(x)−f(y)|≤|f(x)|+|f(y)|<e/2+e/2=e.

So altogether we have ∀x,y∈[a,∞)(|x−y|<d⟹|f(x)−f(y)|<e).

As pointed out in a comment, you're only looking at values near x=a. If you prove that for every ϵ>0 there exists a δ>0 such that |f(x)−f(a)|≤ϵ whenever |x−a|≤δ (which you did not, your proof is confused), then this would only imply the continuity of f at a, not more.

I suggest another approach: note that f is continuous in [a,+∞) (if a>0) and limx→+∞f(x)=0. These two properties combined together give the uniform continuity of f in [a,+∞). The main idea is to combine Heine-Cantor theorem (a continuous function on a compact set is uniformly continuous) with the vanishing condition. Let me know if you manage to conclude or you need a complete proof.